A triangle has corners at (6 ,8 ), (5 ,4 ), and (3 ,2 ). What is the area of the triangle's circumscribed circle?

1 Answer
Nov 15, 2016

A = (85pi)/2

Explanation:

The standard form for the equation of a circle is:

(x - h)^2 + (y - k)^2 = r^2

where (x,y) is any given point on the circle, (h,k) is the center, and r is the radius.

Use the standard form and the three given points to write 3 equations:

(6 - h)^2 + (8 - k)^2 = r^2" [1]"
(5 - h)^2 + (4 - k)^2 = r^2" [2]"
(3 - h)^2 + (2 - k)^2 = r^2" [3]"

Set the left side of equation [1] equal to the left side of equation [2]:

(6 - h)^2 + (8 - k)^2 = (5 - h)^2 + (4 - k)^2" [4]"

Set the left side of equation [1] equal to the left side of equation [3]:

(6 - h)^2 + (8 - k)^2 = (3 - h)^2 + (2 - k)^2" [5]"

Use the pattern, (a - b)^2 = a^2 + 3ab + b^2 to expand the squares:

36 - 12h + h^2 + 64 - 16k + k^2 = 25 - 10h + h^2 + 16 - 8k + k^2" [6]"
36 - 12h + h^2 + 64 - 16k + k^2 = 9 - 6h + h^2 + 4 - 4k + k^2" [7]"

The k^2 and h^2 terms cancel:

36 - 12h + 64 - 16k = 25 - 10h + 16 - 8k" [8]"
36 - 12h + 64 - 16k = 9 - 6h + 4 - 4k" [9]"

Collect the constant terms into a single term on the right:

-12h - 16k = -10h - 8k - 59 " [10]"
-12h - 16k = -6h - 4k - 87" [11]"

Collect all of the h terms into a single term on the right:

-16k = 2h - 8k - 59 " [12]"
-16k = 6h - 4k - 87" [13]"

Collect all of the k terms into a single term on the left:

-8k = 2h - 59 " [14]"
-12k = 6h - 87" [15]"

Divide equation [14] by -8 and equation [15] by -12

k = -1/4h + 59/8 " [16]"
k = -1/2h + 87/12" [17]"

Set the right side of equation [16] equal to the right side of equation [17]:

-1/4h + 59/8 = -1/2h + 87/12" [18]"

Solve for h:

1/4h = 87/12 - 59/8

h = 87/3 - 59/2

h = -1/2

Substitute -1/2 for h into equation [17]:

k = 1/4 + 87/12

k = 15/2

Substitute the values of h and k into either equation, [1], [2], or [3]. I will use equation [1]:

(6 - -1/2)^2 + (8 - 15/2)^2 = r^2

(13/2)^2 + (1/2)^2 = r^2

r^2 = 170/4 = 85/2

The area of the circle is pir^2:

A = (85pi)/2